Calculate the equilibrium constant for the enzymatic hydrolysis of 0.02 M glucose-6-phosphate to glucose and inorganic phosphate given that 0.04 % of the original glucose-6-phosphate remained after reaching equilibrium and the activity of water is unity.

i'm assuming that the correct answer is indeed the first one/ "a"..

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

може ли да зареждате повече информация, моля

5.6⋅10−^19  j

do you need the explaination?